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Conjecture. Let $f:\mathbb R\to\mathbb R_{>0}$ be a continuously differentiable function such that $$f(x+y)\le f(x)\cdot f(y)$$ for all reals $x,y$. Then $$f'(x)\le c_1\cdot\exp(c_2\cdot x)$$ for some constants $c_1,c_2$; and all $x\in\mathbb R$.

Is this conjecture true? How can we (dis-)prove it?


Remarks

  • My conjecture originates from the fact that all continuous functions satisfying $g(x+y)=g(x)\cdot g(y)$ are of the type $g(x)=\exp(c\cdot x)$. In particular, $g'(x)=c\exp(cx).$

  • My first instinct was to use $$f(x+y)-f(x)\le f(x)\cdot(f(y)-1)$$ so if we could conclude that $f(x)\le \exp(cx)$ for all $x$, then $$f'(x)=\lim_{y\to0}\frac{f(x+y)-f(x)}{y}\le \exp(cx)\lim_{y\to0}\frac{\exp(cy)-1}y=\exp(cx)\cdot c.$$ However, $f(x)\le \exp(cx)$ is not always true (consider $f(x)=2\cdot\exp x$).

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This is unfortunately not true. The issue is that your condition essentially only restricts the behaviour of $f$ "in the large", but not locally. The derivative of $f$, however, is determined by the local behaviour. What exactly I mean by this will be made clear by the counterexample.

Define $$ f : R \to (0,\infty), x \mapsto e^x \cdot (4 + \sin(e^{x^{2}})) . $$ Since $-1 \leq \sin(y) \leq 1$, it is then easy to see $3 e^x \leq f(x) \leq 5 \cdot e^x$ for all $x \in \Bbb{R}$, and hence $$ f(x+y) \leq 5 \cdot e^{x+y} \leq 3 e^x \cdot 3 e^y \leq f(x) \cdot f(y). $$ However, we have $$ f'(x) = e^x \cdot (4 + \sin(e^{x^{2}})) + e^x \cdot \cos(e^{x^{2}}) \cdot e^{x^{2}} \cdot 2x. $$ Now, for $n \geq 5$, let $x_n := \sqrt{\ln (2 \pi n)}$. Observe $\ln(2 \pi n) \geq \ln(e) = 1$, and thus $e^{x_n} \geq e \geq 1$. Furthermore, $e^{x_n^2} = e^{\ln(2 \pi n)} = 2\pi n$, and thus $\cos(e^{x_n^2}) = \cos(2\pi n) = 1$. Overall, this implies $$ f'(x_n) \geq 3 \cdot e^{x_n} + e^{x_n} \cdot \cos(e^{x_n^2}) \cdot e^{x_n^2} \cdot 2 x_n \geq 2 \pi n \cdot 2 x_n \geq 4 \pi n. $$ However, if your desired inequality was true, we would have $$ 2 e^{x_n^2} = 4 \pi n \leq f'(x_n) \leq C \cdot e^{c x_n}, $$ and thus $$ e^{x_n^2 - c x_n} \leq C / 2, $$ meaning that $x_n^2 - c x_n = x_n \cdot (x_n - c)$ is bounded, which is clearly not the case.

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    $\begingroup$ I really like your answer. I asked a similar question in the past. Essentially I asked the same thing for a radially symmetric function (if it satisfies the functional inequality, then the gradient grows at most exponentially). Do you believe that one can tweak your example such that it works in my case? The naive approach would be to take your $f$ and to consider $g(x):= f(\vert x\vert)$. But then we need to smooth it at the origin in such a way to it becomes differentiable. This I am unable to do while still preserving the functional inequality. Any idea would be appreciated. $\endgroup$ – Severin Schraven Oct 8 '19 at 20:23
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    $\begingroup$ @Severin: I think $g(x) = f(x) + f(-x)$ should work. I will let you check the details. $\endgroup$ – PhoemueX Oct 8 '19 at 20:31
  • $\begingroup$ Thanks for the input. I'll give it a shot $\endgroup$ – Severin Schraven Oct 8 '19 at 20:33
  • $\begingroup$ @SeverinSchraven My question was inspired by yours 😄 PS: Bis morgen bei den elliptic PDEs 👋 $\endgroup$ – Maximilian Janisch Oct 8 '19 at 20:46
  • $\begingroup$ @MaximilianJanisch Bis morgen! $\endgroup$ – Severin Schraven Oct 8 '19 at 21:00

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